Let $f_5(n)$ be the largest integer $x$ for which $5^x$ divides $n$.
For example, $f_5(625000) = 7$.
Let $T_5(n)$ be the number of integers $i$ which satisfy $f_5((2 \cdot i - 1)!) \lt 2 \cdot f_5(i!)$ and $1 \le i \le n$.
It can be verified that $T_5(10^3) = 68$ and $T_5(10^9) = 2408210$.
Find $T_5(10^{18})$.
Solution
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